# Find the smallest positive number $p$ for which the equation $\cos(p\sin x)=\sin(p \cos x)$ has a solution $x\in[0,2\pi].$

Find the smallest positive number $p$ for which the equation $\cos(p\sin{x})=\sin(p\cos{x})$ has a solution $x$ belonging $[0,2\pi]$. I am not able to solve this problem. Please help me.

It must be that $p\sin{x}+p\cos{x}=\large\frac{\pi}{2}$ with, $p=\frac{\large\frac{\pi}{2}}{\sin{x}+\cos{x}}$. So, to minimize $p$, $\sin{x}+\cos{x}$ must be maximized.

$\sin{x}+\cos{x}=\sqrt2\sin(x+\pi/4)$, which is maximized when $\sin(x+\pi/4)=1$ at $x=\pi/4$.

Hence, $p=\frac{\large\frac{\pi}{2}}{\sqrt2sin(\pi/2)}$

$p=\frac{\pi}{2\sqrt2}$.

$x=\pi/4$.

• the answer in book is pi/2sqrt(2) at x equal to pi/4 and 7pi/4 – Jai Mahajan Jul 30 '14 at 10:20
• Please help me in solving this question. How do we know that this answer is the smallest value. – Jai Mahajan Jul 30 '14 at 10:21
• I guess this solution is more understandable. – Juanito Jul 30 '14 at 19:24

Given two angles $\alpha$ and $\beta$ one has $$\cos\alpha=\sin\beta=\cos(\beta-{\pi\over2})$$ iff either $$\alpha=\beta-{\pi\over2}+2k\pi\tag{1}$$ or $$\alpha=-(\beta-{\pi\over2})+2k\pi,\quad{\rm i.e.}\quad \alpha+\beta={\pi\over2}+2k\pi\ .\tag{2}$$ In our case $\alpha=p\sin x$ and $\beta=p\cos x$, so that $(1)$ is equivalent with $$p(\cos x-\sin x)={\pi\over2}-2k\pi\ ,$$ so that we have to make sure that the equation $$p{2\over\sqrt{2}}\sin({\pi\over4}-x)={\pi\over2}-2k\pi$$ has a real solution $x$. This is the case if $${2\over\sqrt{2}}p \geq|{\pi\over2}-2k\pi|$$ for a suitable $k$, and the smallest positive $p$ that satisfies this is $p={\pi\over 2\sqrt{2}}$. The resulting solution $x$ of the original equation is then $x={7\pi\over4}$.

The case $(2)$ is similar and has the same outcome for $p$ (so that this is the definitive solution of the problem); the corresponding $x$ is then ${\pi\over4}$.